SATHEE: Indefinite Integration 1 Question 35

Indefinite Integration 1 Question 35

36. If $f (x) = A \sin \frac{π x}{2} + B$ ,

$f^{'} \frac{1}{2} = \sqrt{2}$ and $\int_{0}^{1} f (x) d x = \frac{2 A}{π}$ , then constants

$A$ and $B$ are

(1995, 2M)

(a) $\frac{π}{2}$ and $\frac{π}{2}$

(b) $\frac{2}{π}$ and $\frac{3}{π}$

(d) $\frac{4}{π}$ and 0

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Solution:

Given, $f (x) = A \sin \frac{π x}{2} + B, f^{'} \frac{1}{2} = \sqrt{2}$

$and \int_{0}^{1} f (x) d x = \frac{2 A}{π}$

$f^{'} (x) = \frac{A π}{2} \cos \frac{π x}{2} \Rightarrow f^{'} \frac{1}{2} = \frac{A π}{2} \cos \frac{π}{4} = \frac{A π}{2 \sqrt{2}}$

$But f^{'} \frac{1}{2} = \sqrt{2} ∴ \frac{A π}{2 \sqrt{2}} = \sqrt{2} \Rightarrow A = \frac{4}{π}$

Now, $\int_{0}^{1} f (x) d x = \frac{2 A}{π} \Rightarrow \int_{0}^{1} A \sin \frac{π x}{2} + B d x = \frac{2 A}{π}$

$\begin{aligned} \Rightarrow & - \frac{2 A}{π} \cos \frac{π x}{2} + B x & = \frac{2 A}{π} \Rightarrow B + \frac{2 A}{π} = \frac{2 A}{π} \\ \Rightarrow & B & = 0 \end{aligned}$

Indefinite Integration 1 Question 35

36. If $f (x) = A \sin \frac{π x}{2} + B$ ,

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Indefinite Integration 1 Question 35

36. If f(x)=Asin⁡πx2+B,

Solution:

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36. If $f (x) = A \sin \frac{π x}{2} + B$ ,